Theorem 6.4: (The Associative Property for Addition of Natural Numbers) Let a, b, and c be natural numbers. Then

a + (b + c ) = (a + b ) + c.

Proof: By Theorem 6.2 we can find pairwise disjoint sets, R, S, and T, such that

#(R ) = a

#(S ) = b

and

#(T ) = c

Since the sets are pairwise disjoint,

and

Thus

since the union of two empty sets is empty. Thus

by Theorem 2.5, the distributivity of intersections across unions.

Moreover,

and

Thus

again, since the union of two empty sets is empty. Thus

also by Theorem 2.5, the distributivity of intersections across unions. Thus by Theorem 6.1

and

However,

by Theorem 2.3, the associativity of unions of sets, and by Theorem 5.4, the well definedness of cardinality,

(a + b ) + c = a + (b + c )